Question

How many grams of radium can form by passing $\mathbf{235}\mathbf{C}$ through an electrolytic cell containing a molten radium salt?

Short answer

$\mathrm{The}\mathrm{mass}\mathrm{of}\mathrm{radium}\mathrm{produced}\mathrm{is}0.275\mathrm{g}$.

Step-by-step solution

Definition

It's an electrochemical process that involves passing a current between two electrodes through an ionized solution (the electrolyte) to deposit positive ions (anions) on the negative electrode (cathode) and negative ions (cations) on the positive electrode (cathode) (anode).

Calculating the grams of radium

- The charge of $235\mathrm{C}$is passed through an electrolytic cell containing a molten radium salt.

- The charge of radium ion is $2^{+}$, so, the half-reaction for ${\mathrm{Ra}}^{2+}$reduction is

${\mathrm{Ra}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Ra}$

- Faraday constant: charge of a mole of electrons $\left(\mathrm{F}=96485\mathrm{C}/{\mathrm{mol}}^{-}\right)$

- Charge $=235\mathrm{C}$

We can compute the number of moles of electrons using the Faraday constant since we know the charge.

Calculating the mass of radium

$\begin{array}{c}\mathrm{Charge}={\mathrm{Molese}}^{-}\times \mathrm{FMolese}=\frac{\mathrm{Charge}}{\mathrm{F}}\\ =\frac{235\mathrm{C}}{96485\mathrm{C}/{\mathrm{mole}}^{-}}\\ =2.4356\times {10}^{-3}{\mathrm{mole}}^{-}\end{array}$

Because one mole of radium is created by two moles of electrons, the total number of moles of radium produced is

$\mathrm{Moles}\mathrm{Ra}=2.4356\times {10}^{-3}{\mathrm{mole}}^{-}\times \frac{1\mathrm{molRa}}{2{\mathrm{mole}}^{-}}=1.2178\times {10}^{-3}\mathrm{mol}\mathrm{Ra}$

The molar mass of radium is $226\mathrm{g}/\mathrm{mol}$, so, the mass of radium produced is

$\mathrm{Mass}\mathrm{Ra}=1.2178\times {10}^{-3}\mathrm{molRa}\times \frac{226\mathrm{g}}{1\mathrm{mol}\mathrm{Ra}}=0.275\mathrm{g}$

Therefore,$\mathrm{the}\mathrm{mass}\mathrm{of}\mathrm{radium}\mathrm{produced}\mathrm{is}0.275\mathrm{g}$.